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n200080 [17]
3 years ago
12

You are choosing a passcode for your new phone. It has to be a 6 digit code using the numbers 0-9. You are allowed to repeat a n

umber if you would like. What is the probability that someone randomly guesses your combination on the first try?
Mathematics
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

one in a million chance

Step-by-step explanation:

he combination can be 000000 or 999999, or anything in between. Since the odds of getting each digit is 1/10, you multiply 1/10 by itself 6 times. There's literally a one in a million chance of guessing the PIN correctly.

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X is 13 less than 17. WRITE AN EQUATION​ TO REPRESENT THE statement.
vovikov84 [41]

Answer:

X=4

Step-by-step explanation:

13 less than 17 would be 4 becayse if you do 17-13=4

6 0
3 years ago
Given the figure below, which statement correctly completes the following?
vfiekz [6]

Answer:

anyhotty girl uf u bored so zoooomsid 560 571 3546 passwoordd == 123456

5 0
3 years ago
Read 2 more answers
A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic
goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85

Using the above the probabilities compute the remaining values as follows:

P(X^{c})=1-P(X)=1-0.04=0.96

P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10

P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})

          =(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}

             =\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}

             =0.20

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

6 0
3 years ago
Elena receives ​$131 per year in simple interest from three investments totaling ​$3000. Part is invested at​ 3%, part at​ 4% an
ioda

Answer:

Elena invested $ 1,700 at 5%, $ 700 at 4%, and $ 600 at 3%.

Step-by-step explanation:

Given that Elena receives $ 131 per year in simple interest from three investments totaling $ 3000, and part is invested at 3%, part at 4% and part at 5%, and there is $ 1000 more invested at 5% than at 4%, to find the amount invested at each rate, the following calculations must be performed:

1500 x 0.05 + 500 x 0.04 + 1000 x 0.03 = 75 + 20 + 30 = 125

1600 x 0.05 + 600 x 0.04 + 800 x 0.03 = 80 + 24 + 24 = 128

1700 x 0.05 + 700 x 0.04 + 600 x 0.03 = 85 + 28 + 18 = 131

Therefore, Elena invested $ 1,700 at 5%, $ 700 at 4%, and $ 600 at 3%

7 0
3 years ago
Create a pattern with the rule. -4.
Bumek [7]
One pattern is 22, 18, 14, 10, 6, 2 hope this helps just keep subtracting 4. You can go from the highest number to the lowest. Thanks for asking!
7 0
3 years ago
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